${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ Square planar
The electrons are paired in $3\mathrm{d}$ orbitals because ${\mathrm{CN}}^{-}$ is strong field ligand.Hence the hybridisation will be ${\mathrm{dsp}}^2$

$\left[\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\right]$ Tetrahedral
$\mathrm{CO}$ is strong field ligand. First the unpaired electrons will pair and then the electrons from $4\mathrm{s}$ orbital will go in $3\mathrm{d}$ orbital. Now $4\mathrm{s}$ and $4\mathrm{p}$ is empty hence it will be ${\mathrm{sp}}^3$ hybridised.

${\left[{\mathrm{NiCl}}_4\right]}^2$ is paramagnetic in nature due to presence of $2$ unpaired electrons.
${\mathrm{Ni}}^{2+}$ ion has $3{\mathrm{d}}^8$ electronic configuration. It undergoes ${\mathrm{sp}}^3$ hybridisation which results in tetrahedral geometry.